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jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2im confused as to what forces are acting on points a and c.

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2okay, so i did the sum of F in the y direction.. = Ay + Cy  W  T = 0 and then the sum of F in the x direction... = Ax + Cx  Fb = 0 then i did the moment about A and got Fb = 8.9, but i am stuck there.

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2oh and Fb is acting in the negative x direction on point B.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0At point A, it's the Normal reaction from Ground and the Tension from the string... at C it's just the string, is Pulley Masless and frictionless?

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2i assumeee the pulley has no mass nor friction.

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2wait so how many unknowns do i have ?

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0This is getting confusing for me :(

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2me too! haha. and yay '94 kids !

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2i am not sure what direction the force on point a is acting. perpendicular to the rod or the cable?

S
 2 years ago
Best ResponseYou've already chosen the best response.1Does the answer 12.78 lb sound reasonable? I can't remember how do you do this, so I just did a bunch of stuff that seemed to be reasonable and I got that =\

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2it does sound reasonable! but the back of the book has the answers, and evidently T = 9.08 lb ):

S
 2 years ago
Best ResponseYou've already chosen the best response.1Hmmm.. Well at first i did what you did, sum of the forces in the y direction: Ay + Cy  T  15 = 0

S
 2 years ago
Best ResponseYou've already chosen the best response.1A is experiencing a force T, that's pulling on it at 10 degrees, so the y component of that force T that is acting on a would be sin(10) that would be equal to 0.1736T. So now 0.1736T + Cy  1T  15 = 0 0.8264T + Cy  15 = 0

S
 2 years ago
Best ResponseYou've already chosen the best response.1Now if we can substitute something like that for Cy as well we could move on, but idk what to substitute for it cause there are no degrees for point C

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2maybe we are overlooking something. ummmmm.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=cos+40*15%2F%282*cos10*sin40%29

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0\[T \times \cos 10 \times 5 \times \sin 40 = \frac{5}{2} \times \cos 40 \times 15\] T is the tension in the cable. I used concept of Torques, one Torque is in effect due to its weight (rod's weight) the other is from the horizontal component of Cable's Tension (T * cos 10). Note. I calculated Toque about A.

S
 2 years ago
Best ResponseYou've already chosen the best response.1Sum of x direction Ax  Fb = 0 Fb you said = 8.9 Ax  8.9 = 0 Ax = 8.9 cos(10)T = 8.9 0.98T = 8.9 T = 9.08

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2i havent learned torques, but s is something it looks like i can understand. thank you both !!

S
 2 years ago
Best ResponseYou've already chosen the best response.1That kinda looks reasonable when you look at it. I think theres no Cx because its just like a continuation of Ax or something.. idk.. i can't remember.. I took it like a year and a half ago so i can't remember.. i'm gonna be a crappy engineer lol.. can't remember much lol

S
 2 years ago
Best ResponseYou've already chosen the best response.1But at least it give the right answer.. i hope the procedure is right as well =)

jesssica
 2 years ago
Best ResponseYou've already chosen the best response.2sike i still dont know what engineering even is and i am almost done with my first year of college. gracias yall !

S
 2 years ago
Best ResponseYou've already chosen the best response.1Don't worry, you'll figure it out =D De nada =D
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